Case: Meet in the Middle

$C_1 + C_2 = N+1$

When we have a lane with clues on both sides that add up to $N+1$ , we can pinpoint the lane peak.


3			6				4

Take a look at the constraints on where the $6$ can go produced by the $3$ and $4$ clues:


3			6	6	6	6
	6	6	6				4

There’s only 1 spot for the $6$ !

This happens whenever the 2 clues add up to $N+1$ , which is in fact the highest any 2 opposite clues can ever add up to. If they added up to more, the lane would be unsolvable!


4							4
4				6	6	6
	6	6	6				4

Nowhere for $6$ to go!

This deduction is a small reason why closed Skyscrapers are easier than open ones – since every clue has an opposite clue, it’s far more likely you’ll be able to use this deduction to pinpoint a lane peak!